By definition:
For a given function $g(n)$ we denote by $\Theta(g(n))$ the set functions
$\Theta(g(n))$ = $\{f(n):$ there exists positive constants $c_1, c_2$ and $n_0$ such that $0 \leq c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0$ .$\}$
We say $6n^3 \neq \Theta(n^2)$ becuase if it was, then there would be:
\begin{equation} 6n^3 \leq c_2n^2 \Rightarrow n \leq c_2/6 \hspace{100px} (1) \end{equation}
which is not true because $n$ (size of the input) is not limited to any constant.
But to prove $\dfrac{1}{2}n^2-3n = \Theta(n^2)$ we can write
\begin{align*} c_1n^2 &\leq \dfrac{1}{2}n^2-3n \leq c_2n^2\\ c_1 &\leq \dfrac{1}{2}-\dfrac{3}{n} \leq c_2 \end{align*}
by setting $c_1=1/14, c_2=1/2, n_0=7$ it becomes true.
Here is the problem,
$$\dfrac{1}{2}-\dfrac{3}{n} \leq c_2$$
implies
$$ n \leq \dfrac{1/2-c_2}{3} \hspace{100px} (2) $$
why in $(1)$ bounding $n$ is a contradiction, but in $(2)$ it is not?
Thanks